Awire with a weight per unit length of 0.071 n/m is suspended directly above a second wire. the top wire carries a current of 72.8 a, and the bottom wire carries a current of 72.7 a. the permeability of free space is 4π × 10−7 t · m/a . find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. answer in units of mm.

d = 15 mm

Explanation:

Force of repulsion between two current carrying wire is given by

now this force of repulsion is counterbalanced by the weight of the wire

so we have

now we have

so here we can say that

now plug in all values in it

d = 15 mm

magnified images possible images can be real = projected onto a screen

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.